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Math Help - What are the Zero's of this function?

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    What are the Zero's of this function?

    the problem is {x}^{4} - 16

    I get;
    {x}^{4} - {2}^{4} = {(x+2)}^{2} + {(x-2)}^{2}

    for zeros of 2 and -2 each of multiplicity of 2.

    However the book states zeros of 2 and -2 each of multiplicity 1. Could someone be so kind as to explain to me what I have done wrong?
    Last edited by bkbowser; June 1st 2011 at 01:10 PM. Reason: i am bad with latex still
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    Quote Originally Posted by bkbowser View Post
    the problem is {x}^{4} - 16

    I get;
    {x}^{4} - {2}^{4} = {(x+2)}^{2} + {(x-2)}^{2}

    for zeros of 2 and -2 each of multiplicity of 2.

    However the book states zeros of 2 and -2 each of multiplicity 1. Could someone be so kind as to explain to me what I have done wrong?
    You can applied the difference of square incorrectly.

    x^4-16=(x^2)^2-4^2=(x^2-4)(x^2+4)

    Now use the difference of squares again!
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    Quote Originally Posted by TheEmptySet View Post
    You can applied the difference of square incorrectly.

    x^4-16=(x^2)^2-4^2=(x^2-4)(x^2+4)

    Now use the difference of squares again!
    (x^2+4)(x-2)(x+2) Is this it? It does give zeros of 2 and -2 each of multiplicity 1. And since the question just asked about real zeros the other 2 wouldn't count?
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    Quote Originally Posted by bkbowser View Post
    (x^2+4)(x-2)(x+2) Is this it? It does give zeros of 2 and -2 each of multiplicity 1. And since the question just asked about real zeros the other 2 wouldn't count?
    yes
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    Another way to solve x^4- 16= 0 is to write it as x^4= 16. Taking the square root of each side we have x^2= 4 and x^2= -4. From x^2= 4, taking the square root again, x= 2 and x= -2. There are no real number x such that x^2= -4.
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